Ch. 15 & 16 - Acids & Bases

electrolytes
 electrolytes

sour taste

bitter taste

turn litmus red

turn litmus blue

react with metals
to form H2 gas

slippery feel

vinegar, milk, soda,
apples, citrus fruits

ammonia, lye,
antacid, baking soda
ChemASAP

Arrhenius - In aqueous solution…
• Acids form hydronium ions (H3O+)
HCl + H2O 
+
H3O
H
H
Cl
acid
O
H
H
–
+
O
H
+
Cl
H
–
Cl

Arrhenius - In aqueous solution…
• Bases form hydroxide ions (OH-)
NH3 + H2O 
+
NH4
H
H
H
N
H
base
O
H
H
–
+
O
N
H
+
OH
H
H
H

Brønsted-Lowry
• Acids are proton (H+) donors.
• Bases are proton (H+) acceptors.
HCl + H2O 
acid
–
Cl
+
+
H3O
base
conjugate base
conjugate acid
H2O + HNO3  H3O+ + NO3–
B
A
CA
CB
NH3 + H2O 
B
A
 Amphoteric
+
NH4
CA
+
OH
CB
- can be an acid or a base.

Give the conjugate base for each of the following:
HF
F
H3PO4
H2PO4
+
O
H3
 Polyprotic
-
H2O
- an acid with more than one H+

Give the conjugate acid for each of the following:
Br
-
HBr
HSO4
H2SO4
2CO3
HCO3
-

HF + H2O F- +H3O+

CH3COOH + H2O H3O+ + CH3COO -

Lewis
• Acids are electron pair acceptors.
• Bases are electron pair donors.
Lewis
base
Lewis
acid

Strong Acid/Base


HCl
HNO3
H2SO4
HBr
HI
HClO4
100% ionized in water
strong electrolyte
-
+
NaOH
KOH
Ca(OH)2
Ba(OH)2
 Weak
-
HF
CH3COOH
H3PO4
H2CO3
HCN
NH3
Acid/Base
• does not ionize completely
• weak electrolyte
+




Typically double replacement reaction
Hydronium and hydroxide ions react to form
water molecules
Ex.
HCl + NaOH  NaCl + H2O

In an aqueous solution how does the NaOH
dissociate?


How about HCl?


NaOH (aq)  Na+ (aq) + OH- (aq)
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
If you combine the products of the reactions:


Na+ (aq) + OH- (aq) + H3O+ (aq) + Cl- (aq) 
Na+ (aq) + Cl- (aq) +2H2O (aq)
Cancel on both sides (spectator ions):
 OH- (aq) + H3O+ (aq) 2H2O (aq)

Strong Acids completely dissociates to produce
hydronium


Strong electrolyte
Strong bases dissociate to form hydroxide

Strong electrolytes

Refer to Table 15-3 + Table 15-4 on pg 460 - 461for a
list of strong acids and bases. You must know them




Complete the following
1. H2CO3 + Sr(OH)2 
2. HClO4 + NaOH
3. NaHCO3 + H2SO4 

HC2H3O2 + H2O  C2H3O- + H3O+



Refer to table 15-3 in your book, and compare the
strengths of the two acids in the question. Do the
same for the two bases
In which direction, forward or reverse is
favored (it prefers strongest to weakest)
Reverse – production of the weaker acid/ base
is preferred


Pg 478
12, 13,14,15,16,26 ab, 27, 28ab, 30,


As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and
a few act as acids.
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)
 This is referred to as autoionization.



The equilibrium expression for this process is
Kc = [H3O+] [OH−]
This special equilibrium constant is referred to
as the ion-product constant for water, Kw.
At 25°C, Kw = 1.0  10−14M2
- (p)ower of (H)ydrogen
- pH is defined as the negative base-10 logarithm
of the hydronium ion concentration.
pH = −log [H3O+]

In pure water,
Kw = [H3O+] [OH−] = 1.0  10−14

Because in pure water [H3O+] = [OH−],
[H3O+] = (1.0  10−14)1/2 = 1.0  10−7



Therefore, in pure water,
pH = −log (1.0  10−7) = 7.00
An acid has a higher [H3O+] than pure
water, so its pH is <7
A base has a lower [H3O+] than pure water,
so its pH is >7.
These are
the pH
values for
several
common
substances.


The “p” in pH tells us to take the negative log
of the quantity (in this case, hydrogen ions).
Some similar examples are:

pOH = −log [OH−]
Because
[H3O+] [OH−] = Kw = 1.0  10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00

For less accurate
measurements, one
can use

Litmus paper
 “Red” paper turns
blue above ~pH = 8
 “Blue” paper turns
red below ~pH = 5

An indicator
For more accurate
measurements, one
uses a pH meter,
which measures
the voltage in the
solution.



You will recall that the seven strong acids
are HCl, HBr, HI, HNO3, H2SO4, HClO3, and
HClO4.
These are, by definition, strong electrolytes
and exist totally as ions in aqueous solution.
For the monoprotic strong acids,
[H3O+] = [acid].


Strong bases are the soluble hydroxides, which
are the alkali metal and heavier alkaline earth
metal hydroxides (Ca2+, Sr2+, and Ba2+).
Again, these substances dissociate completely
in aqueous solution.

For a generalized acid dissociation,
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
the equilibrium expression would be

[H3O+] [A−]
Kc =
[HA]
This equilibrium constant is called the aciddissociation constant, Ka.
The greater the value of Ka, the stronger
the acid.


The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
We know that
[H3O+] [COO−]
Ka =
[HCOOH]



The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
To calculate Ka, we need the equilibrium
concentrations of all three things.
We can find [H3O+], which is the same as
[HCOO−], from the pH.
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2  10−3 = [H3O+] = [HCOO−]
Now we can set up a table…
[HCOOH], M
Initially
0.10
Change
−4.2  10-3
At
Equilibrium
0.10 − 4.2  10−3
= 0.0958 = 0.10
[H3O+], M [HCOO−], M
0
0
+4.2  10-3 +4.2  10−3
4.2  10−3
4.2  10−3
Ka =
[4.2  10−3] [4.2  10−3]
[0.10]
= 1.8  10−4


[H3O+]eq
Percent Ionization =
 100
[HA]initial
In this example
[H3O+]eq = 4.2  10−3 M
[HCOOH]initial = 0.10 M
4.2  10−3
Percent Ionization =
 100
0.10
= 4.2%
Calculate the pH of a 0.30 M solution of acetic acid,
HC2H3O2, at 25°C.
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)
Ka for acetic acid at 25°C is 1.8  10−5.
The equilibrium constant expression is
[H3O+] [C2H3O2−]
Ka =
[HC2H3O2]
We next set up a table…
[C2H3O2], M
[H3O+], M
[C2H3O2−], M
Initially
0.30
0
0
Change
−x
+x
+x
0.30 − x  0.30
x
x
At Equilibrium
We are assuming that x will be very small compared to 0.30 and
can, therefore, be ignored.
Now,
2
(x)
1.8  10−5 =
(0.30)
(1.8  10−5) (0.30) = x2
5.4  10−6 = x2
2.3  10−3 = x
pH = −log [H3O+]
pH = −log (2.3  10−3)
pH = 2.64


Have more than one acidic proton.
If the difference between the Ka for the first
dissociation and subsequent Ka values is
103 or more, the pH generally depends only
on the first dissociation.
Bases react with water to produce hydroxide
ion.
The equilibrium constant expression for this
reaction is
[HB] [OH−]
Kb =
[B−]
where Kb is the base-dissociation constant.
Kb can be used to find [OH−] and, through it,
pH.
What is the pH of a 0.15 M solution of NH3?
NH3(aq) + H2O(l)
NH4+(aq) + OH−(aq)
[NH4+] [OH−]
Kb =
= 1.8  10−5
[NH3]
Tabulate the data.
[NH3], M
Initially
At Equilibrium
[NH4+], M [OH−], M
0.15
0
0
0.15 - x  0.15
x
x
2
(x)
1.8  10−5 =
(0.15)
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x2
Therefore,
[OH−] = 1.6  10−3 M
pOH = −log (1.6  10−3)
pOH = 2.80
pH = 14.00 − 2.80
pH = 11.20
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can
calculate the other.


Anions are bases.
As such, they can react with water in a
hydrolysis reaction to form OH− and the
conjugate acid:
X−(aq) + H2O(l)
HX(aq) + OH−(aq)


Cations with acidic
protons (like NH4+) will
lower the pH of a solution.
Most metal cations that are
hydrated in solution also
lower the pH of the
solution.



Attraction between nonbonding
electrons on oxygen and the
metal causes a shift of the
electron density in water.
This makes the O-H bond more
polar and the water more acidic.
Greater charge and smaller size
make a cation more acidic.
1.
2.
3.
An anion that is the
conjugate base of a strong
acid will not affect the pH.
An anion that is the
conjugate base of a weak
acid will increase the pH.
A cation that is the
conjugate acid of a weak
base will decrease the pH.
4.
5.
6.
Cations of the strong
Arrhenius bases will not
affect the pH.
Other metal ions will
cause a decrease in pH.
When a solution contains
both the conjugate base of
a weak acid and the
conjugate acid of a weak
base, the affect on pH
depends on the Ka and Kb
values.


The more polar the H-X bond and/or the weaker
the H-X bond, the more acidic the compound.
Acidity increases from left to right across a row
and from top to bottom down a group.
In oxyacids, in
which an OH is
bonded to another
atom, Y, the more
electronegative Y
is, the more acidic
the acid.
For a series of oxyacids, acidity increases
with the number of oxygens.
Resonance in the conjugate bases of
carboxylic acids stabilizes the base and
makes the conjugate acid more acidic.


Lewis acids are defined as electron-pair
acceptors.
Atoms with an empty valence orbital can be
Lewis acids.



Lewis bases are defined as electron-pair donors.
Anything that could be a Brønsted–Lowry base
is a Lewis base.
Lewis bases can interact with things other than
protons, however.